Solutions to 20F Midterm 1, Jan 28, 2009

Little or no partial credit on problems (1), (2), (3E). Seven points for showing T(Q) is nonzero
in problem (4), eight points for the rest of problem (4).

1A)  ..there exist real numbers c_1,...,c_n, not all zero, such that  c_1 v_1 + ...+ c_n v_n = 0.
(The right side is the vector with m zero entries.)

1B) ....of all linear combinations of v_1,..., v_n.

2A)  For each  y in R^m , y = T(x) for some x in R^n.

2B)  For each pair  u, v in R^n,  T(u) = T(v)  implies  u=v.

3A)  After two row switches, A becomes

1  0  1  0  2
2  1  5  1  7
0  0  0  1  2  .

Subtract twice row 1 from row 2 to get

1  0  1  0  2
0  1  3  1  3
0  0  0  1  2  .

This is in echelon form--to get reduced echelon form, subtract row 3 from row 2 to get U.

3B)  The solutions vectors x to  Ax = 0 are the same as the solutions x  to Ux = 0.
Consider  x = (x1, x2, x3, x4, x5) as a column vector, for ease of notation.
The pivot variables are x1, x2, x4, and the free variables are x3 and x5.
Solve for the pivot variables in terms of the free variables.
x5 = x5, x4 = -2x5, x3 = x3, x2 = -3x3 - x5, x1 = -x3 -2x5.
Thus the column vectors c and d are  c = (-1, -3, 1, 0, 0) and d = (-2, -1, 0, -2, 1).

3C)  Repeat the procedure in (3A) for the augmented matrix [A|b] (instead of just for A) to turn it into [U|q],
where q is the column vector (1,6, -3).  Solve as in (3B) to see that the set of all solutions to Ax=b 
is  p + Span{c,d}, where the column vector p is a "particular solution" (1, 6, 0, -3, 0).

3D)  Yes. Since A has 3 pivot variables, Ax = v has a solution x for every vector v in R^3.
Thus every v in R^3 is a linear combination of the columns of A.

3E) The matrix in question is U^T A.  The bottom row of U^T is (2,1,2), and the rightmost column of A
is (2,2,7).  The dot product of these two vectors gives the answer,  20.

4)  The line L has the parametric equation  P + t Q,  where the parameter t is real.  (Line L is parallel
to the vector Q, and line L passes through the point P.)
By linearity of T,  we have T(L) = {T(P+tQ):  t in R} = {T(P) + tT(Q): t in R}.
This is a line through the point T(P) parallel to the vector T(Q), provided T(Q) is nonzero.
To see that T(Q) is nonzero, we need the one-to-one hypothesis.  Since  T(0)=0 by linearity
(e.g., pick c=0 in the formula T(cv) = cT(v)), we cannot also have T(Q)=0, otherwise
the two distinct vectors 0 and Q would  both have the same image, violating the one-to-one hypothesis.